斯特拉森矩阵乘法的实现

网上关于斯特拉森矩阵的乘法，代码框架很多

只是没有实现的代码块（有，但是就是不会运行）

老师布置的了相关利用分治法来做的的题目

想了很久，大概三天了（分治法大问题化小问题）是真没有完美的解决方法

找到了个可以运行的java代码：java写的代码思路是真的清晰（粉java）

package sort;

/**
 * @Author liguo
 * @Description
 * @Data 2018-04-07 14:44
 */
public class Sitelasen {

    public static void main(String[] args) {
        int[][] cc = {{1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1}};
        int[][] e = new Sitelasen().multiMatrix( cc, cc );
        for (int i = 0; i < e.length; i++) { 
            for (int j = 0; j < e.length; j++)
                System.out.print( e[i][j] + " " );
            System.out.println();
        }
    }

    int[][] multiMatrix(int[][] x, int[][] y)//矩阵的乘法
    {
        Sitelasen matr = new Sitelasen();
        int a = x.length / 2;
        if (x.length == 2) {
            int m1 = (x[0][1] - x[1][1]) * (y[1][0] + y[1][1]),
                    m2 = (x[0][0] + x[1][1]) * (y[0][0] + y[0][1]),
                    m3 = (x[1][0] - x[0][0]) * (y[0][0] + y[0][1]),
                    m4 = (x[0][0] + x[0][1]) * y[1][1],
                    m5 = x[0][0] * (y[0][1] - y[1][1]),
                    m6 = x[1][1] * (y[1][0] - y[0][0]),
                    m7 = (x[1][0] + x[1][1]) * y[0][0];
            int[][] c = new int[2][2];
            c[0][0] = m1 + m2 + m6 - m4;
            c[0][1] = m4 + m5;
            c[1][0] = m6 + m7;
            c[1][1] = m2 + m3 + m5 - m7;
            return c;
        } else {
            int[][] a11 = new Sitelasen().sepaMatr( x, 0, 0 );
            int[][] a12 = new Sitelasen().sepaMatr( x, 0, a );
            int[][] a21 = new Sitelasen().sepaMatr( x, a, 0 );
            int[][] a22 = new Sitelasen().sepaMatr( x, a, a );
            int[][] b11 = new Sitelasen().sepaMatr( y, 0, 0 );
            int[][] b12 = new Sitelasen().sepaMatr( y, 0, a );
            int[][] b21 = new Sitelasen().sepaMatr( y, a, 0 );
            int[][] b22 = new Sitelasen().sepaMatr( y, a, a );
            int[][] m1 = matr.multiMatrix( matr.operaMatr( a11, a22, '-' ), matr.operaMatr( b21, b22, '+' ) ),
                    m2 = matr.multiMatrix( matr.operaMatr( a11, a22, '+' ), matr.operaMatr( b11, b22, '+' ) ),
                    m3 = matr.multiMatrix( matr.operaMatr( a21, a11, '-' ), matr.operaMatr( b11, b12, '+' ) ),
                    m4 = matr.multiMatrix( matr.operaMatr( a12, a11, '+' ), b22 ),
                    m5 = matr.multiMatrix( a11, matr.operaMatr( b12, b22, '-' ) ),
                    m6 = matr.multiMatrix( a22, matr.operaMatr( b21, b11, '-' ) ),
                    m7 = matr.multiMatrix( matr.operaMatr( a21, a22, '+' ), b11 );
            int[][] c11 = matr.operaMatr( matr.operaMatr( m6, m4, '-' ), matr.operaMatr( m1, m2, '+' ), '+' );
            int[][] c12 = matr.operaMatr( m4, m5, '+' );
            int[][] c21 = matr.operaMatr( m6, m7, '+' );
            int[][] c22 = matr.operaMatr( matr.operaMatr( m2, m3, '+' ), matr.operaMatr( m5, m7, '-' ), '+' );
            return matr.mergMatr( c11, c12, c21, c22 );

        }
    }

    int[][] mergMatr(int[][] x11, int[][] x12, int[][] x21, int[][] x22)//和并矩阵，将c11,c12,c21,c22合并成一个矩阵
    {
        int[][] a = new int[2 * x11.length][2 * x11.length];
        for (int i = 0; i < x11.length; i++)
            for (int j = 0; j < x11.length; j++) {
                a[i][j] = x11[i][j];
            }
        for (int i = 0; i < x11.length; i++)
            for (int j = 0; j < x11.length; j++) {
                a[i][j + x11.length] = x12[i][j];
            }
        for (int i = 0; i < x11.length; i++)
            for (int j = 0; j < x11.length; j++) {
                a[i + x11.length][j] = x21[i][j];
            }
        for (int i = 0; i < x11.length; i++)
            for (int j = 0; j < x11.length; j++) {
                a[i + x11.length][j + x11.length] = x22[i][j];
            }
        return a;
    }

    int[][] sepaMatr(int[][] x, int m, int n)//分解矩阵成矩阵块
    {
        int[][] a = new int[x.length / 2][x.length / 2];
        for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a.length; j++) {
                a[i][j] = x[m + i][n + j];
            }
        return a;
    }

    int[][] operaMatr(int[][] x, int[][] y, char f)//矩阵的加减
    {
        if (f == '+') {
            int[][] a = new int[x.length][x.length];
            for (int i = 0; i < x.length; i++)
                for (int j = 0; j < x.length; j++)
                    a[i][j] = x[i][j] + y[i][j];
            return a;
        } else if (f == '-') {
            int[][] a = new int[x.length][x.length];
            for (int i = 0; i < x.length; i++)
                for (int j = 0; j < x.length; j++)
                    a[i][j] = x[i][j] - y[i][j];
            return a;
        } else {
            System.out.println( "error" );
            int[][] a = new int[1][1];
            return a;
        }
    }

}
 

 同时也有了一个可以跑起来的c++实现：

#include<iostream>

#include<math.h>

#include<fstream>

using namespace std;

ifstream infile("123.txt",ios::in);

void Input(int n,int **A)

{

    //infile>>n;

    for(int i=0;i<n;i++)

    for(int j=0;j<n;j++)

    infile>>A[i][j];

}

void Output(int n,int **A)

{

    for(int i=0;i<n;i++)

{

    for(int j=0;j<n;j++)

    cout<<A[i][j]<<'	';

    cout<<endl;

}

cout<<endl;

 

}

void Divide(int n,int **A,int **A11,int **A12,int **A21,int **A22)

{

    int i,j;

    for(i=0;i<n;i++)

    for(j=0;j<n;j++)

    {

        A11[i][j]=A[i][j];

        A12[i][j]=A[i][j+n];

        A21[i][j]=A[i+n][j];

        A22[i][j]=A[i+n][j+n];

    }

 

}

void Unit(int n,int **A,int **A11,int **A12,int **A21,int **A22)

{

    int i,j;

    for(i=0;i<n;i++)

    for(j=0;j<n;j++)

    {

        A[i][j]=A11[i][j];

        A[i][j+n]=A12[i][j];

        A[i+n][j]=A21[i][j];

        A[i+n][j+n]=A22[i][j];

    }

}

void Sub(int n,int **A,int **B,int **C)

{

    int i,j;

    for(i=0;i<n;i++)

    for(j=0;j<n;j++)

    C[i][j]=A[i][j]-B[i][j];

}

void Add(int n,int **A,int **B,int **C)

{

    int i,j;

    for(i=0;i<n;i++)

    for(j=0;j<n;j++)

    C[i][j]=A[i][j]+B[i][j];

}

void Mul(int n,int **A,int **B,int **M)

{

    if(n==1)

    M[0][0]=A[0][0]*B[0][0];

    else

    {

    n=n/2;

    int **A11,**A12,**A21,**A22;

    int **B11,**B12,**B21,**B22;

    int **M11,**M12,**M21,**M22;

    int **M1,**M2,**M3,**M4,**M5,**M6,**M7;

    int **T1,**T2;

    A11=new int*[n];

    A12=new int*[n];

    A21=new int*[n];

    A22=new int*[n];

 

    B11=new int*[n];

    B12=new int*[n];

    B21=new int*[n];

    B22=new int*[n];

 

        M11=new int*[n];

        M12=new int*[n];

        M21=new int*[n];

        M22=new int*[n];

 

    M1=new int*[n];

    M2=new int*[n];

    M3=new int*[n];

    M4=new int*[n];

    M5=new int*[n];

    M6=new int*[n];

    M7=new int*[n];

 

    T1=new int*[n];

    T2=new int*[n];

 

        int i;

    for(i=0;i<n;i++)

    {

        A11[i]=new int[n];

        A12[i]=new int[n];

        A21[i]=new int[n];

        A22[i]=new int[n];

        B11[i]=new int[n];

        B12[i]=new int[n];

        B21[i]=new int[n];

        B22[i]=new int[n];

        M11[i]=new int[n];

        M12[i]=new int[n];

        M21[i]=new int[n];

        M22[i]=new int[n];

        M1[i]=new int[n];

        M2[i]=new int[n];

        M3[i]=new int[n];

        M4[i]=new int[n];

        M5[i]=new int[n];

        M6[i]=new int[n];

        M7[i]=new int[n];

 

        T1[i]=new int[n];

        T2[i]=new int[n];

 

    }

    Divide(n,A,A11,A12,A21,A22);

    Divide(n,B,B11,B12,B21,B22);

   // cout<<"A11,A12,A21,A22"<<endl;

   // Output(n,A11);Output(n,A12);Output(n,A21);Output(n,A22);

        Sub(n,B12,B22,T1);

     //   cout<<"B12-B22"<<endl;

      //  Output(n,T1);

        Mul(n,A11,T1,M1);

 

        Add(n,A11,A12,T2);

        Mul(n,T2,B22,M2);

 

        Add(n,A21,A22,T1);

        Mul(n,T1,B11,M3);

 

        Sub(n,B21,B11,T1);

        Mul(n,A22,T1,M4);

 

        Add(n,A11,A22,T1);

        Add(n,B11,B22,T2);

        Mul(n,T1,T2,M5);

 

        Sub(n,A12,A22,T1);

        Add(n,B21,B22,T2);

        Mul(n,T1,T2,M6);

 

        Sub(n,A11,A21,T1);

        Add(n,B11,B12,T2);

        Mul(n,T1,T2,M7);

 

 

        Add(n,M5,M4,T1);

        Sub(n,T1,M2,T2);

        Add(n,T2,M6,M11);

 

        Add(n,M1,M2,M12);

 

        Add(n,M3,M4,M21);

 

        Add(n,M5,M1,T1);

        Sub(n,T1,M3,T2);

        Sub(n,T2,M7,M22);

 

        Unit(n,M,M11,M12,M21,M22);

 

 

 

    }

}

 

int main()

{

    int n;

    cout<<"please input number n"<<endl;

    cin>>n;

    int **A,**B,**C;

    A=new int*[n];

    B=new int*[n];

    C=new int*[n];

    for(int i=0;i<n;i++)

    {

        A[i]=new int[n];

        B[i]=new int[n];

        C[i]=new int[n];

    }

    Input(n,A);

    cout<<"A Matrix is"<<endl;

    Output(n,A);

    Input(n,B);

    cout<<"B Matrix is"<<endl;

    Output(n,B);

    Input(n,C);

    //Output(n,C);

    Mul(n,A,B,C);

    cout<<"The Product of A and B is"<<endl;

    Output(n,C);

   // cout<<n<<endl;

    infile.close();

    return 0;

}

//就编程实来讲，以后我再也不会花费大量时间来训练类似的题目了

实现是一方面，更多的是你在历尽很多bug后，还是没有做出来能够跑起来的程序就很无奈

//本来自己还想做任意阶矩阵乘法的，实现不了，就降低了难度，想着做n*n阶矩阵，到最后只想做出来二的整数次幂阶的斯特拉森矩阵乘法。